The area (in sq.units) of the region bounded | vedclass

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(C) The given parabolas are $y^2 = 4x$ and $y^2 = 4(4-x)$.To find the intersection points,set $4x = 4(4-x)$,which gives $x = 4-x$,so $2x = 4$,implying

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$\frac{32\sqrt{2}}{3}$

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(C) The given parabolas are $y^2 = 4x$ and $y^2 = 4(4-x)$.To find the intersection points,set $4x = 4(4-x)$,which gives $x = 4-x$,so $2x = 4$,implying $x = 2$.At $x = 2$,$y^2 = 4(2) = 8$,so $y = \pm 2\sqrt{2}$.The area is symmetric about the $x$-axis,so we calculate the area in the first quadrant and multiply by $2$.Area $= 2 \int_{0}^{2} \sqrt{4x} \, dx + 2 \int_{2}^{4} \sqrt{4(4-x)} \, dx$.Area $= 2 	imes 2 \int_{0}^{2} \sqrt{x} \, dx + 2 	imes 2 \int_{2}^{4} \sqrt{4-x} \, dx$.Area $= 4 \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{2} + 4 \left[ \frac{-(4-x)^{3/2}}{3/2} ight]_{2}^{4}$.Area $= 4 	imes \frac{2}{3} [2^{3/2} - 0] + 4 	imes \frac{2}{3} [-(0) - (-(4-2)^{3/2})]$.Area $= \frac{8}{3} [2\sqrt{2}] + \frac{8}{3} [2\sqrt{2}] = \frac{16\sqrt{2}}{3} + \frac{16\sqrt{2}}{3} = \frac{32\sqrt{2}}{3}$.

The area (in sq.units) of the region bounded by the parabolas $y^2=4x$ and $y^2=4(4-x)$ is

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